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                <div class="post-meta-line"><span class="post-author"><a href="/" title="Author" rel=" author" class="author"><i class="fas fa-user-circle fa-fw"></i>作者</a></span>&nbsp;<span class="post-category">出版于  <a href="/categories/%E7%AE%97%E6%B3%95%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3/"><i class="far fa-folder fa-fw"></i>算法——滑动窗口</a></span></div>
                <div class="post-meta-line"><span><i class="far fa-calendar-alt fa-fw"></i>&nbsp;<time datetime="2022-10-07">2022-10-07</time></span>&nbsp;<span><i class="fas fa-pencil-alt fa-fw"></i>&nbsp;约 975 字</span>&nbsp;
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                </div><div class="content" id="content"><div class="highlight"><pre tabindex="0" class="chroma"><code class="language-cpp" data-lang="cpp"><span class="k">class</span> <span class="nc">Solution</span> <span class="p">{</span>
 <span class="k">public</span><span class="o">:</span>
	<span class="k">const</span> <span class="kt">int</span> <span class="n">mod</span> <span class="o">=</span> <span class="mf">1e9</span><span class="o">+</span><span class="mi">7</span><span class="p">;</span>
	<span class="kt">int</span> <span class="nf">beautifulBouquet</span><span class="p">(</span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;&amp;</span> <span class="n">flowers</span><span class="p">,</span> <span class="kt">int</span> <span class="n">cnt</span><span class="p">)</span> <span class="p">{</span>

		<span class="kt">int</span> <span class="n">n</span> <span class="o">=</span> <span class="n">flowers</span><span class="p">.</span><span class="n">size</span><span class="p">();</span>
		<span class="kt">int</span> <span class="n">ret</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span>
		<span class="kt">int</span> <span class="n">left</span> <span class="o">=</span> <span class="mi">0</span><span class="p">,</span><span class="n">right</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span>
		<span class="n">unordered_map</span><span class="o">&lt;</span><span class="kt">int</span><span class="p">,</span><span class="kt">int</span><span class="o">&gt;</span> <span class="n">count</span><span class="p">;</span>
		<span class="kt">int</span> <span class="n">option</span> <span class="o">=</span> <span class="mi">1</span><span class="p">;</span>

		<span class="k">while</span><span class="p">(</span><span class="n">right</span> <span class="o">&lt;=</span> <span class="n">n</span><span class="p">){</span>
			<span class="k">if</span><span class="p">(</span><span class="n">option</span> <span class="o">==</span> <span class="mi">1</span><span class="p">){</span> <span class="c1">//right指针向右边滑动，更新窗口上限
</span><span class="c1"></span>				<span class="k">if</span><span class="p">(</span><span class="n">right</span> <span class="o">==</span> <span class="n">n</span><span class="p">){</span>
					<span class="n">option</span> <span class="o">=</span> <span class="mi">2</span><span class="p">;</span>
					<span class="k">continue</span> <span class="p">;</span>
				<span class="p">}</span>
				<span class="kt">int</span> <span class="n">p</span> <span class="o">=</span> <span class="o">++</span><span class="n">count</span><span class="p">[</span><span class="n">flowers</span><span class="p">[</span><span class="n">right</span><span class="o">++</span><span class="p">]];</span>
				<span class="k">if</span><span class="p">(</span><span class="n">p</span> <span class="o">&gt;</span> <span class="n">cnt</span><span class="p">){</span>
					<span class="n">option</span> <span class="o">=</span> <span class="mi">2</span><span class="p">;</span>
				<span class="p">}</span>
			<span class="p">}</span><span class="k">else</span><span class="p">{</span> <span class="c1">//left指针滑动开始收缩窗口，同时不断更新答案
</span><span class="c1"></span>				<span class="kt">int</span> <span class="n">span</span> <span class="o">=</span>  <span class="n">right</span> <span class="o">-</span><span class="mi">1</span> <span class="o">-</span><span class="n">left</span><span class="p">;</span>
				<span class="k">if</span><span class="p">(</span> <span class="n">right</span> <span class="o">==</span> <span class="n">n</span><span class="o">&amp;&amp;</span> <span class="n">count</span><span class="p">[</span><span class="n">flowers</span><span class="p">.</span><span class="n">back</span><span class="p">()]</span> <span class="o">&lt;=</span> <span class="n">cnt</span><span class="p">){</span> <span class="c1">//2
</span><span class="c1"></span>					<span class="n">span</span> <span class="o">=</span> <span class="n">right</span> <span class="o">-</span> <span class="n">left</span><span class="p">;</span>
				<span class="p">}</span>
				<span class="n">ret</span> <span class="o">=</span> <span class="p">(</span><span class="n">ret</span> <span class="o">+</span> <span class="n">span</span><span class="p">)</span> <span class="o">%</span> <span class="n">mod</span><span class="p">;</span>
				<span class="kt">int</span> <span class="n">p</span> <span class="o">=</span> <span class="o">--</span><span class="n">count</span><span class="p">[</span><span class="n">flowers</span><span class="p">[</span><span class="n">left</span><span class="o">++</span><span class="p">]];</span>
				<span class="k">if</span><span class="p">(</span><span class="n">p</span> <span class="o">==</span> <span class="n">cnt</span><span class="p">){</span>
					<span class="n">option</span> <span class="o">=</span> <span class="mi">1</span><span class="p">;</span>
				<span class="p">}</span>
			<span class="p">}</span>
			<span class="k">if</span><span class="p">(</span><span class="n">left</span> <span class="o">==</span> <span class="n">right</span><span class="p">)</span>
				<span class="k">break</span> <span class="p">;</span>
		<span class="p">}</span>

		<span class="k">return</span> <span class="n">ret</span><span class="p">;</span>
	<span class="p">}</span>
<span class="p">};</span>
</code></pre></div><p><a href="https://leetcode.cn/problems/1GxJYY/" target="_blank" rel="noopener noreffer">题目链接</a></p>
<p>这道题首先由简单的枚举来过渡：比如1232，cnt=1。</p>
<p>我们普通的做法可以这样枚举：从左到右枚举首元素：以1开头有1、12、123，以2开头有2、23，以3开头有3、32，以2开头有2.</p>
<p>很明显这样枚举是可行的，但是需要O(n^2)的时间复杂度，外层for枚举开头，里层寻找终点（连续不包含cnt个重复数字的最长序列）。</p>
<p>比如我最开始就是这样写的：很自然的就超时了</p>
<div class="highlight"><pre tabindex="0" class="chroma"><code class="language-cpp" data-lang="cpp"><span class="k">class</span> <span class="nc">Solution</span> <span class="p">{</span>
<span class="k">public</span><span class="o">:</span>
    <span class="k">const</span> <span class="kt">int</span> <span class="n">mod</span> <span class="o">=</span> <span class="mf">1e9</span><span class="o">+</span><span class="mi">7</span><span class="p">;</span>
    <span class="kt">int</span> <span class="nf">beautifulBouquet</span><span class="p">(</span><span class="n">vector</span><span class="o">&lt;</span><span class="kt">int</span><span class="o">&gt;&amp;</span> <span class="n">flowers</span><span class="p">,</span> <span class="kt">int</span> <span class="n">cnt</span><span class="p">)</span> <span class="p">{</span>
        <span class="kt">int</span> <span class="n">n</span> <span class="o">=</span> <span class="n">flowers</span><span class="p">.</span><span class="n">size</span><span class="p">();</span>
        <span class="kt">int</span> <span class="n">ret</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span>
        <span class="k">for</span><span class="p">(</span><span class="kt">int</span> <span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">i</span><span class="o">&lt;</span><span class="n">n</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">){</span>
            <span class="kt">int</span> <span class="n">ma</span> <span class="o">=</span> <span class="mi">0</span><span class="p">;</span>
            <span class="n">unordered_map</span><span class="o">&lt;</span><span class="kt">int</span><span class="p">,</span><span class="kt">int</span><span class="o">&gt;</span> <span class="n">mmp</span><span class="p">;</span>
            <span class="kt">int</span> <span class="n">j</span><span class="o">=</span><span class="n">i</span><span class="p">;</span>
            <span class="k">for</span><span class="p">(;</span><span class="n">j</span><span class="o">&lt;</span><span class="n">n</span><span class="p">;</span><span class="n">j</span><span class="o">++</span><span class="p">){</span>
                <span class="kt">int</span> <span class="n">p</span> <span class="o">=</span> <span class="o">++</span><span class="n">mmp</span><span class="p">[</span><span class="n">flowers</span><span class="p">[</span><span class="n">j</span><span class="p">]];</span>
                <span class="n">ma</span> <span class="o">=</span> <span class="n">max</span><span class="p">(</span><span class="n">ma</span><span class="p">,</span><span class="n">p</span><span class="p">);</span>
                <span class="k">if</span><span class="p">(</span><span class="n">ma</span> <span class="o">&gt;</span> <span class="n">cnt</span><span class="p">){</span>
                    <span class="k">break</span><span class="p">;</span>
                <span class="p">}</span>
            <span class="p">}</span>
            <span class="n">ret</span> <span class="o">=</span> <span class="p">(</span><span class="n">ret</span><span class="o">+</span><span class="p">(</span><span class="n">j</span><span class="o">-</span><span class="n">i</span><span class="p">))</span><span class="o">%</span><span class="n">mod</span><span class="p">;</span>
        <span class="p">}</span>
        <span class="k">return</span> <span class="n">ret</span><span class="p">;</span>
    <span class="p">}</span>
<span class="p">};</span>
</code></pre></div><p>我们在枚举里层循环的时候，发现有多次重复计算，比如1开头的尾部是3，2开头的尾部还是3，那么该怎么优化呢？直接做缓存处理也是不可行的，因为无法确定后续遍历的右边界都一致。</p>
<p>那么如何优化呢？很快我们可以想到建立一个窗口来滑动的思想，我们的问题在于无法确认当前遍历到的头部的右边界（重复数字）是否还是前一个数确认的边界，滑动窗口可以通过哈希表记录简单的解决这一问题。</p>
<ol>
<li>
<p>我们先把窗口向右扩张(<code>right++、++count[x]</code>)，也就是随便先找到一个右边界（这个边界内重复的数字不超过 <code>cnt</code> 个）。</p>
</li>
<li>
<p>不断的收缩窗口并更新答案(<code>left--,--count[x]</code>)，如果在收缩过程中 <code>count[x]</code> 被减小到正好等于 <code>cnt</code> ，那就说明之后 <code>left</code> 经过元素的有边界发生了改变，此时需要再次向右扩大窗口找到新的边界。</p>
</li>
</ol>
<p>1和2的过程不断循环重复，直到 <code>left</code> 遍历并更新完了所有的元素，即 <code>left == right</code> 时，跳出循环。</p>
<blockquote>
<p>小细节：有两种情况需要被分清楚否则会出错，1.right往后扩散发现没有重复元素（即此时没有了右边界） 2.右边界的位置就是最后一个元素的位置，即right的位置在最后一个元素的下标位置.</p>
<p>在我写的代码中，由于left和right始终指向还未扫描过的位置，所以上述两种情况right的值会相同都为flowers数组的大小n。但这两种情况计算答案则不同，第二种情况会少一种情况。所以有了上述题解代码的 <code>2</code> 进行判断和处理。</p>
</blockquote>
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